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# Why can't Sp3 hybrid orbitals can form pi bonds?

A **single bond** consists of a ##sigma## bond, which is defined as the **head-on overlap** between two compatible orbitals.

Because hybrid orbitals are made to match the symmetry of the **incoming** (i.e. not-yet-bonded) atom's atomic orbital for the sake of being able to bond at all (which is usually if not always "totally symmetric about the internuclear axis"), they **always** have to overlap head-on.

Thus, a hybrid orbital has to make a ##sigma## bond.

Not just ##sp^3##, but **any** hybrid orbital. Even in a **triple bond**, like in acetylene (##"H"-"C"-="C"-"H"##), the ##pi## bonds are made by the ##p_x## and ##p_y## orbitals (or any qualified equivalent sidelong orbital overlap), while the ##sigma## bonds are made with the hybrid orbitals, which consist of only the ##p_z## and ##s## orbitals.

Here, the ##z## axis is defined as the internuclear axis, whereas the ##x## and ##y## axes are vertical and towards us, respectively.

As an alternative, consider ethene (##"H"_2"C"="CH"_2##), where the double bonds lie on the ##xy##-plane in the plane of the screen/paper, and the ##z## axis protrudes outward towards us.

If we consider the ##sp^2## hybridization of carbon in ethene, carbon actually uses **three** ##sp^2## hybrid orbitals: **one each** to ##sigma## bond with the hydrogens, and **one** to ##sigma## bond with the other carbon. (You would get that if you drew the molecule with all single bonds and no double bonds.)

It made each **hybrid orbital** by **mixing** the ##2s##, ##2p_x##, and ##2p_y## orbitals together and distributing their angular positions evenly on the ##xy##-plane since the original ##2p## orbitals used to construct them were actually aligned in the ##x## and ##y## directions, respectively. (The notation ##sp^2## comes from mixing one ##s## and two ##p## orbitals together to get ##33%## ##s## character and ##66%## ##p## character.)

**Hence, the** ##\mathbf(sigma)## **bonds of ethene (shown in yellow) lie on a single plane**.

In addition, we have the ##\mathbf(2p_z)## **atomic orbital** (perpendicular to the plane of the molecule, in purple) from each carbon that is **used to** ##\mathbf(pi)## **bond** (sidelong overlap) with the other carbon, thus constructing the second component of the double bond (one ##sigma## + one ##pi## bond = one double bond).

It was **not used** in the orbital hybridization, and it remains as a **different, incompatible orbital (with respect to the** ##\mathbf(2p_x)## **and** ##\mathbf(2p_y)##**) for** ##\mathbf(sigma)## **bonding** within the molecule. The only thing it can do at this point is ##pi## bond because it is oriented precisely to do so.